`vec(r)=15t^(2)i+(20-20t^(2))j`
find magnitude of acceleration at `t=1` sec.

Options

A. `30m//s^(2)`
B. `40m//s^(2)`
C. `70m//s^(2)`
D. `50m//s^(2)`

Answer

To find the magnitude of the acceleration at \( t = 1 \) second for the given position vector \( \vec{r} = 15t^2 \hat{i} + (20 - 20t^2) \hat{j} \), we will follow these steps:

### Step 1: Differentiate the position vector to find the velocity vector.
The position vector is given as:
\[ \vec{r} = 15t^2 \hat{i} + (20 - 20t^2) \hat{j} \]

To find the velocity vector \( \vec{v} \), we differentiate \( \vec{r} \) with respect to time \( t \):
\[ \vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(15t^2) \hat{i} + \frac{d}{dt}(20 - 20t^2) \hat{j} \]

Calculating the derivatives:
- For the \( \hat{i} \) component:
\[ \frac{d}{dt}(15t^2) = 30t \]
- For the \( \hat{j} \) component:
\[ \frac{d}{dt}(20 - 20t^2) = 0 - 40t = -40t \]

Thus, the velocity vector is:
\[ \vec{v} = 30t \hat{i} - 40t \hat{j} \]

### Step 2: Differentiate the velocity vector to find the acceleration vector.
Next, we find the acceleration vector \( \vec{a} \) by differentiating the velocity vector:
\[ \vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(30t) \hat{i} + \frac{d}{dt}(-40t) \hat{j} \]

Calculating the derivatives:
- For the \( \hat{i} \) component:
\[ \frac{d}{dt}(30t) = 30 \]
- For the \( \hat{j} \) component:
\[ \frac{d}{dt}(-40t) = -40 \]

Thus, the acceleration vector is:
\[ \vec{a} = 30 \hat{i} - 40 \hat{j} \]

### Step 3: Calculate the magnitude of the acceleration vector.
The magnitude of the acceleration vector \( |\vec{a}| \) is given by:
\[ |\vec{a}| = \sqrt{(30)^2 + (-40)^2} \]

Calculating:
\[ |\vec{a}| = \sqrt{900 + 1600} = \sqrt{2500} = 50 \, \text{m/s}^2 \]

### Conclusion
The magnitude of the acceleration at \( t = 1 \) second is:
\[ \boxed{50 \, \text{m/s}^2} \]